Path Sum

　　题目链接

　　题目要求：

　　Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

　　For example:
　　Given the below binary tree and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1

　　return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

　　这道题采用深度优先搜索的方法就可以了，具体程序（12ms）如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if(!root)
14             return false;
15         return hasPathSumSub(root, 0, sum);
16     }
17
18     bool hasPathSumSub(TreeNode *tree, int subSum, int sum)
19     {
20         if(!tree)
21             return false;
22
23         subSum += tree->val;
24         if(!tree->left && !tree->right && subSum == sum)
25             return true;
26
27         return hasPathSumSub(tree->left, subSum, sum) || hasPathSumSub(tree->right, subSum, sum);
28     }
29 };

Path Sum II

　　题目链接

　　题目要求：

　　Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.

　　For example:
　　Given the below binary tree and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \    /         7    2  5   1

　　return

[
   [5,4,11,2],
   [5,8,4,5]
]

　　这题与上题类似。具体程序（24ms）如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> pathSum(TreeNode* root, int sum) {
13         vector<vector<int>> rVec;
14         if(!root)
15             return rVec;
16
17         vector<int> vec;
18         hasPathSumSub(root, vec, rVec, 0, sum);
19         return rVec;
20     }
21
22     void hasPathSumSub(TreeNode *tree, vector<int> vec, vector<vector<int>>& rVec, int subSum, int sum)
23     {
24         if(!tree)
25             return;
26
27         vec.push_back(tree->val);
28         subSum += tree->val;
29         if(!tree->left && !tree->right && subSum == sum)
30             rVec.push_back(vec);
31
32         hasPathSumSub(tree->left, vec, rVec, subSum, sum);
33         hasPathSumSub(tree->right, vec, rVec, subSum, sum);
34     }
35 };